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Some quick calc questions

  1. Sep 17, 2005 #1
    ok, first what is e^t times e^-t?
    and can anyone help with this one:

    the integral from 0 to pi of (sinti+costj+tk)?
    is it -costi + sintj +(t^2/2)k?

    thank you!
  2. jcsd
  3. Sep 17, 2005 #2
    Well e^-t x e^t is same as e^t-t so It should be e^0=1.

    I havent started on Integrals yet.
  4. Sep 17, 2005 #3
    I believe

    [tex] e^t e^{ - t} = e^0 = 1 [/tex]


    [tex]\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2}[/tex]
    :smile: Just remember to take the limits of the integral.
    You have antiderivative, now apply the algebra
    Last edited: Sep 18, 2005
  5. Sep 18, 2005 #4


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    Homework Helper

    bomba923, I think you've made a mistake here, :wink:
    First, I think the question reads:
    [tex]\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt[/tex]
    And second, where does the little 'C' come from? :smile:
    Nope, it isn't.
    In your post, you didn't state the variable of integration. So I assume it is dt.
    [tex]\int (\sin (it) + \cos (jt) + kt) dt = \int \sin (it) dt + \int \cos (jt) dt + \int kt dt[/tex]
    [tex]= \int \frac{1}{i} \sin (it) d(it) + \int \frac{1}{j} \cos (jt) d(jt) + k \int t dt[/tex]
    [tex]= \frac{1}{i} \int \sin (it) d(it) + \frac{1}{j} \int \cos (jt) d(jt) + k \int t dt = ...[/tex]
    Can you go from here?
    Viet Dao,
    Last edited: Sep 18, 2005
  6. Sep 18, 2005 #5
    You have confused me here.

    I thought the question asked:
    [tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]

    My answer is
    [tex] i + j + \frac{{\pi ^2 }}{2}k[/tex]
  7. Sep 18, 2005 #6


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    Homework Helper

    If I am asking some question like that, I am going to put the i, and j in front of the cos and sin functions.
    Something like this:
    i sint + j cost + kt.

    And even if the question reads like this:
    [tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]
    That is still not the correct answer. :wink:
    If the question happens to be:
    int (i sint + j cost + kt) dt
    You can use bomba923's suggestion. You should omit the constant of integration 'C' in his answer.
    Viet Dao,
    Last edited: Sep 18, 2005
  8. Sep 18, 2005 #7
    Ok, but I learnt to put i, j and k at the end when I did vectors....

    I did it again and got the same answer. Where does 2i come from?

    How stupid of me, I had pi/2 in my head instead of pi.
    Last edited: Sep 18, 2005
  9. Sep 18, 2005 #8


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    Science Advisor

    Certainly anyone taking calculus should have learned long ago that axbx= ax+y. In particular, axa-x= a0= 1. The fact that a= e is irrelevant.

    Yes, the original post's "integral from 0 to pi of (sinti+costj+tk)" should have been written
    [tex]\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k} dt[/itex]
    to distinguish it form [tex]\int_0^{\pi}{sin ti+ cos tj+ tk} dt[/itex] but I think the "i, j, k" suggested vectors rather than numbers.

    don23, It is true that
    [tex]\int {(sin t)i+ (cos t)j+ (t)k}dt= -cos(t)i+ sin(t)j+ (\frac{1}{2}t^2+ C[/tex]
    but you forgot to evaluate it between 0 and [itex]\pi[/itex].

    [tex](-cos \pi)i+ (sin /pi)j+ \frac{1}{2}(/pi)^2= i+ 0j+ \frac{\pi^2}{2}k[/tex]
    [tex](-cos 0)i+ (sin 0)i+ \frac{1}{2}(0)^2= -i+ 0j+ 0k[/tex]
    [tex]\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k}dt= 2i +\frac{\pi^2}{2}k[/tex]
  10. Sep 18, 2005 #9
    Thank you Halls of Ivy :redface:
    (you've vindicated me!)
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